POJ 3214 Object Clustering 哈夫曼距离最小生成树

题目：http://poj.org/problem?id=3241

题意：平面中给定n个点，任意两点之间的距离为它们的哈夫曼距离，求n个点的最小生成树中的第k大边

思路：kuangbin大神的模板，看了好久。。。这个不错：点这里

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef long long ll;
const int N = 100010;
const int INF = 0x3f3f3f3f;
struct node
{
    int x, y, id;
}p[N];
struct edge
{
    int v, u, d;
}g[N];
struct BIT
{
    int minn, pos;
    void init()
    {
        minn = INF, pos = -1;
    }
}bit[N];
int n, k, tot;
int par[N];
bool cmp(node a, node b)
{
    if(a.x != b.x) return a.x < b.x;
    else return a.y < b.y;
}
int ask(int i, int m)
{
    int minn = INF, pos = -1;
    while(i <= m)
    {
        if(bit[i].minn < minn)
            minn = bit[i].minn, pos = bit[i].pos;
        i += i & -i;
    }
    return pos;
}
void update(int i, int val, int pos)
{
    while(i > 0)
    {
        if(val < bit[i].minn)
            bit[i].minn = val, bit[i].pos = pos;
        i -= i & -i;
    }
}
void add_edge(int v, int u, int d)
{
    g[tot].v = v, g[tot].u = u, g[tot++].d = d;
}
void mmst(int n, node p[])
{
    int a[N], b[N];
    tot = 0;
    for(int i = 0; i < 4; i++)
    {
        if(i == 1 || i == 3)
        {
            for(int j = 0; j < n; j++)
                swap(p[j].x, p[j].y);
        }
        else if(i == 2)
        {
            for(int j = 0; j < n; j++)
                p[j].x = -p[j].x;
        }
        sort(p, p + n, cmp);
        for(int j = 0; j < n; j++)
            a[j] = b[j] = p[j].y - p[j].x;
        sort(b, b + n);
        int m = unique(b, b + n) - b;
        for(int j = 1; j <= m; j++)
            bit[j].init();
        for(int j = n - 1; j >= 0; j--)
        {
            int pos = lower_bound(b, b + m, a[j]) - b + 1;//按照y-x离散数据
            int ans = ask(pos, m); //查询[pos, m]中x+y最小值的编号
            if(ans != -1)
                add_edge(p[j].id, p[ans].id, abs(p[j].x - p[ans].x) + abs(p[j].y - p[ans].y));
            update(pos, p[j].x + p[j].y, j); //更新
        }
    }
}
bool cmpg(edge a, edge b)
{
    return a.d < b.d;
}
int ser(int x) //并查集查询
{
    int r = x, i = x, j;
    while(r != par[r]) r = par[r];
    while(par[i] != r) j = par[i], par[i] = r, i = j;
    return r;
}
int solve(int k)
{
    mmst(n, p); //预处理，曼哈顿距离最小生成树
    for(int i = 0; i < n; i++) par[i] = i; //并查集初始化
    sort(g, g + tot, cmpg); //按边的权值排序
    for(int i = 0; i < tot; i++) //kruskal算法求最小生成树，因为从小到大，所以求到第n-k小的边就是答案
    {
        int fv = ser(g[i].v), fu = ser(g[i].u);
        if(fv != fu)
        {
            par[fu] = fv;
            if(--k == 0) return g[i].d;
        }
    }
}
int main()
{
    while(~ scanf("%d%d", &n, &k))
    {
        for(int i = 0; i < n; i++)
            scanf("%d%d", &p[i].x, &p[i].y), p[i].id = i;
        printf("%d\n", solve(n - k));
    }
    return 0;
}